PropositionProperties of matrix inversion

The inverse operation on matrices has the following properties

  1. If the \((n,n)\)-matrix \(\Mtrx{A}\) is invertible, then an inverse of \(\Mtrx{A}\) has size \((n,n)\).

  2. If a matrix is invertible, then it has exactly one inverse.

  3. If \(\Mtrx{A}\) is invertible, then \(A^{-1}\) is also invertible, and   \((\Mtrx{A}^{-1})^{-1} = \Mtrx{A}\).

  4. If \(\Mtrx{A}\) and \(\Mtrx{B}\) are invertible matrices, then \(\Mtrx{A}\Mtrx{B}\) is also invertible, and   \((\Mtrx{A}\Mtrx{B})^{-1} = \Mtrx{B}^{-1} \Mtrx{A}^{-1}\).

Proof

i. Size of the inverse

Suppose \(\Mtrx{B}\) is an inverse of \(\Mtrx{A}\). Then the identity \(\Mtrx{A}\Mtrx{B} = \IdMtrx{n}\) implies that the product \(\Mtrx{A}\Mtrx{B}\) is defined, and so \(\Mtrx{B}\) has \(n\) rows. Moreover, \(\Mtrx{B}\) has \(n\) columns because \(\IdMtrx{n}\) has \(n\) columns.

ii. Uniqueness of the inverse

Suppose the square matrix \(\Mtrx{A}\) is invertible, and that we have two inverses \(\Mtrx{B}\) and \(\Mtrx{C}\). We need to show that \(\Mtrx{B} = \Mtrx{C}\). - Indeed, we know that

\(\Mtrx{A} \Mtrx{B}\)\(= \IdMtrx{n} =\)\(\Mtrx{B} \Mtrx{A}\)
\(\Mtrx{A} \Mtrx{C}\)\(= \IdMtrx{n} =\)\(\Mtrx{C} \Mtrx{A}\)

But then we have the following computation

\(\Mtrx{A}\Mtrx{B}\)\(=\)\(\Mtrx{A}\Mtrx{C}\)
\(\Mtrx{B}(\Mtrx{A}\Mtrx{B})\)\(=\)\(\Mtrx{B}(\Mtrx{A}\Mtrx{C})\)
\((\Mtrx{B}\Mtrx{A})\Mtrx{B}\)\(=\)\((\Mtrx{B}\Mtrx{A})\Mtrx{C}\)
\(\IdMtrx{n}\Mtrx{B}\)\(=\)\(\IdMtrx{n}\Mtrx{C}\)
\(\Mtrx{B}\)\(=\)\(\Mtrx{C}\)

and this is exactly what we wanted to show.

iii.

Since \(\Mtrx{A}\) is invertible, we know that there is a matrix \(\Mtrx{B}\) with

\[\Mtrx{A}\Mtrx{B} = \IdMtrx{n} = \Mtrx{B}\Mtrx{A}\]

Further, we know that this \(\Mtrx{B}\) is unique. So we set \(\Mtrx{A}^{-1} \DefEq \Mtrx{B}\). Now the identity above can also be interpreted as saying that \(\Mtrx{B}\) is invertible and that its unique inverse is \(\Mtrx{A}\). Therefore

\[\Mtrx{A} = \Mtrx{B}^{-1} = (\Mtrx{A}^{-1})^{-1},\]

which is exactly what we wanted to show.

iv. Inverse of a product

If \(\Mtrx{A}\) and \(\Mtrx{B}\) are invertible matrices, we need to show that

\[(\Mtrx{A}\Mtrx{B})(\Mtrx{B}^{-1}\Mtrx{A}^{-1}) = \IdMtrx{n} = (\Mtrx{B}^{-1}\Mtrx{A}^{-1})(\Mtrx{A}\Mtrx{B})\]

To see that the left hand side identity is true, we compute

\((\Mtrx{A}\Mtrx{B})(\Mtrx{B}^{-1}\Mtrx{A}^{-1})\)\(= \)\(\Mtrx{A} (\Mtrx{B}\Mtrx{B}^{-1}) \Mtrx{A}^{-1}\)
\(\)\(= \)\(\Mtrx{A} \IdMtrx{n} \Mtrx{A}^{-1}\)
\(\)\(= \)\(\Mtrx{A} \Mtrx{A}^{-1}\)
\(\)\(=\)\(\IdMtrx{n}\)

The identity on the right follows from a similar argument.